{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "This notebook was prepared by [hashhar](https://github.com/hashhar), second solution added by [janhak] (https://github.com/janhak). Source and license info is on [GitHub](https://github.com/donnemartin/interactive-coding-challenges)."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Solution Notebook"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Problem: Compress a string such that 'AAABCCDDDD' becomes 'A3BCCD4'.  Only compress the string if it saves space.\n",
    "\n",
    "* [Constraints](#Constraints)\n",
    "* [Test Cases](#Test-Cases)\n",
    "* [Algorithm](#Algorithm)\n",
    "* [Code](#Code)\n",
    "* [Unit Test](#Unit-Test)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Constraints\n",
    "\n",
    "* Can we assume the string is ASCII?\n",
    "    * Yes\n",
    "    * Note: Unicode strings could require special handling depending on your language\n",
    "* Is this case sensitive?\n",
    "    * Yes\n",
    "* Can we use additional data structures?  \n",
    "    * Yes\n",
    "* Can we assume this fits in memory?\n",
    "    * Yes"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Test Cases\n",
    "\n",
    "* None -> None\n",
    "* '' -> ''\n",
    "* 'AABBCC' -> 'AABBCC'\n",
    "* 'AAABCCDDDD' -> 'A3BCCD4'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Algorithm\n",
    "\n",
    "Since Python strings are immutable, we'll use a list of characters to build the compressed string representation.  We'll then convert the list to a string.\n",
    "\n",
    "* Calculate the size of the compressed string\n",
    "    * Note the constraint about compressing only if it saves space\n",
    "* If the compressed string size is >= string size, return string\n",
    "* Create compressed_string\n",
    "    * For each char in string\n",
    "        * If char is the same as last_char, increment count\n",
    "        * Else\n",
    "            * If the count is more than 2\n",
    "                * Append last_char to compressed_string\n",
    "                * append count to compressed_string\n",
    "                * count = 1\n",
    "                * last_char = char\n",
    "            * If count is 1\n",
    "                * Append last_char to compressed_string\n",
    "                * count = 1\n",
    "                * last_char = char\n",
    "            * If count is 2\n",
    "                * Append last_char to compressed_string\n",
    "                * Append last_char to compressed_string once more\n",
    "                * count = 1\n",
    "                * last_char = char\n",
    "        * Append last_char to compressed_string\n",
    "        * Append count to compressed_string\n",
    "    * Return compressed_string\n",
    "\n",
    "Complexity:\n",
    "* Time: O(n)\n",
    "* Space: O(n)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Code"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": false
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   "outputs": [],
   "source": [
    "def compress_string(string):\n",
    "    if string is None or len(string) == 0:\n",
    "        return string\n",
    "\n",
    "    # Calculate the size of the compressed string\n",
    "    size = 0\n",
    "    last_char = string[0]\n",
    "    for char in string:\n",
    "        if char != last_char:\n",
    "            size += 2\n",
    "            last_char = char\n",
    "    size += 2\n",
    "\n",
    "    # If the compressed string size is greater than\n",
    "    # or equal to string size, return original string\n",
    "    if size >= len(string):\n",
    "        return string\n",
    "\n",
    "    # Create compressed_string\n",
    "    # New objective:\n",
    "    # Single characters are to be left as is\n",
    "    # Double characters are to be left as are\n",
    "    compressed_string = list()\n",
    "    count = 0\n",
    "    last_char = string[0]\n",
    "    for char in string:\n",
    "        if char == last_char:\n",
    "            count += 1\n",
    "        else:\n",
    "            # Do the old compression tricks only if count exceeds two\n",
    "            if count > 2:\n",
    "                compressed_string.append(last_char)\n",
    "                compressed_string.append(str(count))\n",
    "                count = 1\n",
    "                last_char = char\n",
    "            # If count is either 1 or 2\n",
    "            else:\n",
    "                # If count is 1, leave the char as is\n",
    "                if count == 1:\n",
    "                    compressed_string.append(last_char)\n",
    "                    count = 1\n",
    "                    last_char = char\n",
    "                # If count is 2, append the character twice\n",
    "                else:\n",
    "                    compressed_string.append(last_char)\n",
    "                    compressed_string.append(last_char)\n",
    "                    count = 1\n",
    "                    last_char = char\n",
    "    compressed_string.append(last_char)\n",
    "    compressed_string.append(str(count))\n",
    "\n",
    "    # Convert the characters in the list to a string\n",
    "    return \"\".join(compressed_string)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Algorithm: Split to blocks and compress\n",
    "\n",
    "Let us split the string first into blocks of identical characters and then compress it block by block.\n",
    "\n",
    "* Split the string to blocks\n",
    "    * For each character in string\n",
    "        * Add this character to block\n",
    "        * If the next character is different\n",
    "            * Return block\n",
    "            * Erase the content of block\n",
    "\n",
    "\n",
    "* Compress block\n",
    "    *  If block consists of two or fewer characters\n",
    "        * Return block\n",
    "    * Else\n",
    "        * Append length of the block to the first character and return\n",
    "\n",
    "\n",
    "* Compress string\n",
    "    * Split the string to blocks\n",
    "    * Compress blocks\n",
    "    * Join compressed blocks\n",
    "    * Return result if it is shorter than original string\n",
    "\n",
    "Complexity:\n",
    "* Time: O(n)\n",
    "* Space: O(n)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "def split_to_blocks(string):\n",
    "    block = ''\n",
    "    for char, next_char in zip(string, string[1:] + ' '):\n",
    "        block += char\n",
    "        if char is not next_char:\n",
    "            yield block\n",
    "            block = ''\n",
    "\n",
    "\n",
    "def compress_block(block):\n",
    "    if len(block) <= 2:\n",
    "        return block\n",
    "    else:\n",
    "        return block[0] + str(len(block))\n",
    "\n",
    "\n",
    "def compress_string(string):\n",
    "    if string is None or not string:\n",
    "        return string\n",
    "    compressed = (compress_block(block) for block in split_to_blocks(string))\n",
    "    result = ''.join(compressed)\n",
    "    return result if len(result) < len(string) else string"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Unit Test"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": false
   },
   "outputs": [],
   "source": [
    "%%writefile test_compress.py\n",
    "from nose.tools import assert_equal\n",
    "\n",
    "\n",
    "class TestCompress(object):\n",
    "\n",
    "    def test_compress(self, func):\n",
    "        assert_equal(func(None), None)\n",
    "        assert_equal(func(''), '')\n",
    "        assert_equal(func('AABBCC'), 'AABBCC')\n",
    "        assert_equal(func('AAABCCDDDD'), 'A3BCCD4')\n",
    "        assert_equal(func('aaBCCEFFFFKKMMMMMMP taaammanlaarrrr seeeeeeeeek tooo'), 'aaBCCEF4KKM6P ta3mmanlaar4 se9k to3')\n",
    "        print('Success: test_compress')\n",
    "\n",
    "\n",
    "def main():\n",
    "    test = TestCompress()\n",
    "    test.test_compress(compress_string)\n",
    "\n",
    "\n",
    "if __name__ == '__main__':\n",
    "    main()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": false
   },
   "outputs": [],
   "source": [
    "%run -i test_compress.py"
   ]
  }
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